Question
Evaluate the following integrals:
$\int\sin^3\text{x}\cos^5\text{x}\text{ dx}$
$\int\sin^3\text{x}\cos^5\text{x}\text{ dx}$
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Answer
$\int\sin^3\text{x}\cos^5\text{x}\text{ dx}$
$=\int\sin^2\text{x}\cdot\cos^5\text{x}\cdot\sin\text{x}\text{ dx}$
$=\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x}\text{ dx}=\text{dt}$
$\sin\text{x}\text{ dx}=-\text{dt}$
Now, $\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
$=-\int(1-\text{t}^2)\text{t}^5\text{dt}$
$=-(\text{t}^5-\text{t}^7)\text{dt}$
$=-\int(\text{t}^7-\text{t}^5)\text{dt}$
$=\frac{\text{t}^8}{8}-\frac{\text{t}^6}{6}+\text{C}$
$=\frac{\cos^8\text{x}}{8}-\frac{\cos^6\text{6}\text{x}}{6}+\text{C}$
$=\int\sin^2\text{x}\cdot\cos^5\text{x}\cdot\sin\text{x}\text{ dx}$
$=\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
Let $\cos\text{x}=\text{t}$
$-\sin\text{x}\text{ dx}=\text{dt}$
$\sin\text{x}\text{ dx}=-\text{dt}$
Now, $\int(1-\cos^2\text{x})\cdot\cos^5\text{x}\sin\text{x}\text{ dx}$
$=-\int(1-\text{t}^2)\text{t}^5\text{dt}$
$=-(\text{t}^5-\text{t}^7)\text{dt}$
$=-\int(\text{t}^7-\text{t}^5)\text{dt}$
$=\frac{\text{t}^8}{8}-\frac{\text{t}^6}{6}+\text{C}$
$=\frac{\cos^8\text{x}}{8}-\frac{\cos^6\text{6}\text{x}}{6}+\text{C}$
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